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## Homework Statement

A patrol boat off a coast is ordered to destroy a rocket launcher that sits on the roof of 40-meter tall building. The horizontal distance between the boat and the rocket launcher is 1500 meters. The muzzle velocity of the patrol boat’s stabilized 30-mm cannon is 141 meters/second. What are the two angles (relative to the horizontal) at which the patrol boat’s gun can be aimed in order to destroy the rocket launcher? Express your answer in degrees to a precision of three significant figures. [You may ignore air resistance. Assume an acceleration of gravity of precisely 9.800 m/s2. Your answers must be accurate to within 0.3 degrees in order to receive credit. You must also show your work.]

Vo= 141 m/s

X= 1500 m

Yf = 40 m

Yo = 0

t = ?

Theta = ?

## Homework Equations

Yf=Yo + Voy(t) - .5(9.8)(tsquared)

## The Attempt at a Solution

One angle seems to be 23.4 degrees but I cant understand how you would find the other angle. I divided the initial velocity by the x distance to find 10.34 seconds. Plugged that into the Yf=Yo + Voy(t) - .5(9.8)(tsquared) equation with Yf being 40 meters, Yo being zero and Voy being (Vo)(sin)(theta) and then solving for theta. I checked it and it is accurate within a meter to the final position of y being 40 meters so the math is accurate even if my theory may be incorrect. However, I dont see any strategy for finding the second angle.

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